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For a signed measure $\mu$ on the Borel $\sigma$-algebra of $\mathbb{R}$ satisfying $\vert \mu \vert < \infty$ is it always possible to find a sequence of measures $\{\mu_n\}$, each a linear combination of Dirac measures, converging weakly to $\mu$? By weakly I mean

$$\int f(x) \mu_n(\text{d} x) \rightarrow \int f(x) \mu(\text{d} x)$$

as $n \rightarrow \infty$ for every bounded, continuous function $f$. If not, does it help to restrict the assumptions, e.g. to positive measures?

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  • $\begingroup$ I think that the answer is affirmative as a consequence of Krein-Milman's theorem: try looking in the book by Simon. $\endgroup$ – Giuseppe Negro Apr 27 '18 at 10:05
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Yes. First, since $||\mu||<\infty$ and $f$ is bounded, $$\int_{[-n,n]}f\,d\mu\to\int f\,d\mu.$$So you can assume that $\mu$ has compact support.

Now if $\mu$ is supported in $[-A,A]$ then $f$ is uniformly continuous on $[-A-1,A+1]$, hence $$\mu_n=\sum_j\mu([j/n,(j+1)/n))\delta_{j/n}$$works.

Edit: It's been pointed out that I'm implicitly swapping two limits in saying that we can assume $\mu$ has compact support. Above I say more or less that $\chi_{[-n,n]}d\mu\to d\mu$ weakly; in fact this convergence is in norm, and that saves the day:

For any measure $\nu$ define $$\nu_n=\sum_{j=-n^2}^{n^2}\nu([j/n,(j+1)/n))\delta_{j/n}.$$

If $\mu$ is a real measure then $\mu_n\to\mu$ weakly: Fix $\epsilon>0$. Define $$d\nu^A=\chi_{[-A,A]}d\nu.$$ If $A$ is large enough then $||\mu-\mu^A||<\epsilon$ and also $||\mu_n^A-\mu_n||<\epsilon$ for every $n$. Fix such an $A$. (Oops, the notation is ambiguous: By $\mu_n^A$ I mean $(\mu^A)_n$.)

As above we have $\lim_{n\to\infty}\int fd\mu_n^A=\int fd\mu^A$. So the triangle inequality shows that $$\left|\int fd\mu_n-\int fd\mu\right|<\epsilon(1+2||f||_\infty)$$if $n$ is large enough.

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    $\begingroup$ Thanks for the answer. I just realized, however, that I don't quite follow one part of your construction. Suppose the measure does not have compact support, then we can divide $\mathbb{R}$ into the sets $[-n,n]$ as you suggest and on each of these intervals the measures, call then $\{\mu_m\}$, will converge. If I have understood you correctly, we are left with the situation $$\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\int_{[-n,n]}f\text{d}\mu_m = \int f \text{d}\mu,$$ but we would like the limits to be in the reverse order. How can we justify changing the order of the limits? $\endgroup$ – Billy Pilgrim Apr 29 '18 at 12:02
  • $\begingroup$ Fair point. In fact we can swap those limits because $\int_{[-n,n]}fd\mu\to\int fd\mu$ uniformly for $||f||\le1$; see edit for details. $\endgroup$ – David C. Ullrich Apr 29 '18 at 13:31

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